If it cannot, is it because the definition of [tex]\frac{dy}{dx}[/tex] that says that [tex]\frac{dy}{dx}=\lim_{\Delta x\rightarrow 0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}[/tex] requires that [tex]dx[/tex] must be the independent variable and [tex]dy[/tex] must be the dependent variable? Or, is there any other reason?

Actually, in most mathematics there is little or no distinction between "independent" and "dependent" variables. All we really need is that there be a "relation" between x and y. If y= sin(x), as long as we stay on an interval in which sine is "one-to-one", we can define the inverse function x= arcsin(y). Another way to find the derivative of x= sin(y), with respect to x, is to use "implicit differentiation": take the derivative of both sides with respect to x. 1= cos(y)dy/dx (by the chain rule) so dy/dx= 1/cos(y)= sec(y). If it bothers you that the right side is a function of y instead of x, you can change that by using trig identities, or even simpler thinking of y as an angle in a right triangle. If x= sin(y), then we can take x as the "opposite side" of a right triangle with hypotenuse 1. y is the angle opposite side x of course, so sin(y)= opposite side/near side= x/1= x. By the Pythagorean Theorem, then, the length of the "near side" is [itex]\sqrt{1- x^2}[/itex]. Since secant is defined as "hypotenuse over near side", sec(y)= [itex]1/\sqr{1- x^2}[/itex]. That is, if x= sin(y), dx/dy= [itex]1/\sqrt{1- x^2}[/itex] which is precisely the derivative formula for "arcsin(x)".